- --- In primenumbers@yahoogroups.com, Wed Mar 29, 2006, "Jens Kruse

Andersen" <jens.k.a@...> wrote:> You may have misunderstood this "limit".

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> If c = 2, 3, 5, 11, 17, or 41, then x^2-x+c is prime for x = 0..c-1.

> It has been proved there are no other such c values.

> See http://mathworld.wolfram.com/LuckyNumberofEuler.html

> (That page incorrectly says x = 0..c-2 but it makes no

> difference for the solutions).

> However, there may be large c for which x^2-x+c is prime

> for x = 0..41 or longer. Finding such a c is infeasible.

Thank you, Jens, for this contribution.

I would like to give here some comments :

1. There is effectively an error in this mathworld's page.

I am going to write a short e-mail to the team of Eric W. Weisstein.

The polynomials x^2 - x + c are often sources of confusion, and it

could be also the case here about this definition of the lucky numbers

of Euler. We don't know if the author wanted to write x = 0 ... c-1

(because he wanted all the primes, repeated or not) or x = 1 ... c-1

(because he wanted only the different prime numbers and didn't care to

begin with x = 1) or x = 0 ... c-2 (because he wanted the same range

as for the Euler-like polynomials x^2 + x + c).

But there is another possibility, more plausible : this page correctly

says that " n = 0 ... p-2 ", but the author put wrongly n^2 - n + p

instead of n^2 + n + p, whereas he put this last Euler-like

polynomial in the following link :

http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

Note also the following prime curio given by Le Lionnais for the

number 41 (see http://primes.utm.edu/curios/page.php ) : it's one of

the numbers p (or "Lucky Numbers of Euler") such that a Euler-like

polynomial is prime for n = 0, 1, ..., p-2.

Anyway, the lucky numbers of Euler are concerning the two families of

polynomials, even if the reference to the mathematician Euler in the

definition should privilege the Euler-like polynomials n^2 + n + p.

Do they have to be mentioned both in the definition of the lucky

numbers of Euler ?

2. Your idea concerning this "limit" is interesting.

In the message 17837 (about polynomials of the form (m.x - k)^2 + (m.x

- k) + p ) i wrote this : " My contribution was not concerning

quadratic polynomials givning less than p-1 distinct primes. An open

problem is to find out polynomials of degree 2 giving L distinct

primes when p > 41 and 40 < L < p-1 ".

But what you are saying here, about polynomials of the form ax^2 - bx

+ c, is more general. You consider that the "limitation" of the lucky

numbers of Euler will not concern the other quadratic polynomials (of

the form ax^2 -bx + c) with more than 40 (i.e., the highest lucky

number of Euler - 1) distinct positive prime numbers.

So, by transforming my sentence above and using the polynomials ax^2 +

bx+ c (to avoid any risk of confusion with ax^2 - bx + c), we could

simply say : " an open problem is to find out polynomials of degree 2

giving L distinct primes, with L > 40 ". In a first approach, this

situation seems to be more general than mine : starting with your

interpretation and the polynomials ax^2 + bx + c, with c = p prime, p> 41, L > 40 and x = 0 ... L-1, we have three possibilities :

A. L < p-1

B. L = p-1

C. L > p-1

The case A corresponds to my description above.

The case B is impossible because p will never be a lucky number of

Euler when p > 41.

The case C paradoxically brings us to the case B : if a polynomial

ax^2 + bx + p, with p > 41, gives more than p-1 distinct prime

numbers, it gives already distinct prime numbers when we only consider

x = 0 ... p-2.

The interest of your proposal, Jens, is finally to confirm that the

only possible polynomials to be sought are the polynomials included in

A, where ax^2 + bx + p gives L distinct positive primes when p > 41

and 40 < L < p-1.

3. As i wrote above, the polynomials x^2 - x + c are often sources of

problems. But the differences between x^2 - x + c and x^2 + x + c

(when c is a lucky number of Euler) are not summarized with the fact

that they are both giving p-1 distinct positive primes in a different

range of the variable x (x = 1 ... c-1 for the first family

and x = 0 ... c-2 for the second family).

Something else appears when we compare them ...

I would like to do a general remark on the nature of the independent

term c when we look at general polynomials ax^2 + bx + c (with integer

coefficients) giving any number of distinct positive prime numbers.

We know that if we want a sequence of distinct and positive prime

numbers, such that this sequence begins with x = 0, then we require

that c is positive and prime. If it is not the case, then we have to

change our way of presentation and begin all the sequences with x = 1

(which is more logical in the sense that the FIRST value of x gives

the FIRST value of the sequence). By the only fact that we are

starting with x = 0, we artificially create a constraint on the nature

of c, which must be prime in any case (i mean in all the polynomials

giving any number of distinct positive primes). We do not have

necessarily such a constraint when we decide that all the sequences

must be counted from x = 1, and in this case we could imagine that c

is not systematically a prime number.

For instance, when i discussed about the polynomials (m.x - k)^2 +

(m.x - k) + c, with m > 0 and k = 0, c-1, c-2, 2c-3 (see

http://groups.yahoo.com/group/primenumbers/message/17841 ), i was

obliged to consider that c is prime in each case of production of

sequences of primes. Only because we require to start with w = 0. So

all the families of polynomials giving p-1 prime numbers present

consequently a prime independent term (a lucky number of Euler or

'derivated' from it). Hence, the search for polynomials giving more

than 40 positive distinct numbers could be strongly influenced by the

initial conditions of counting.

I don't know if this situation needs to do an adaptation of the

definition of the prime-generating polynomials (perhaps the best is to

continue not to precise how we are counting) or to be more interested

in polynomials ax^2 - bx + c with an only aim of widening the

possibilities for c (anyway i am not sure that they will give more

good polynomials than ax^2 + bx +c).

The attitude depends finally on what we are searching for : sequences

with repeated prime numbers or ... or sequences with distinct positive

prime numbers. For this last kind of search for instance, the

polynomials x^2 - x + c , with c prime, are excluded when we count

from x = 0 because they give a repeated prime number in the begin of

the sequence.

In conclusion, i would say that by starting the sequences at x = 1, it

is not excluded that we enlarge the field of possibilities and give

more chance to find out quadratic polynomials giving more than 40

positive distinct prime numbers.

Patrick Capelle. - --- In primenumbers@yahoogroups.com, Wed March 29, 2006, "Jens Kruse

Andersen" <jens.k.a@...> wrote:> Heuristics support this plausible conjecture:

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> For any n>0 and k, there is a polynomial of degree n which

> gives distinct positive primes for the first k integers.

The zoo of prime numbers did not accustom us to so much simplicity and

absence of special constraints.

In spite of similar formal aspects, we are far here from the idea that

for any k, there is a polynomial of degree n which gives distinct

positive primes for the first k integers.

Paradoxical situation : we have a lot of difficulties to find out one

quadratic polynomial giving more than 40 distinct positive prime

numbers , and during this time one can imagine the existence of a

quadratic polynomial P(x) giving distinct and positive prime numbers

when x goes from 0 to 10^10 ...

Regards,

Patrick Capelle. - Only a comment.

If you want to prove a polynomials P(x) is prime,

isnt enough to determine the assoiated roots.

E.g. P(x)=x^2+2

The number N is P(x) which can base expanded in

some base b as

N=\sum_{i=0}^m c_i b^i .

Determining the roots and the coefficients I claim

is order \ln^3(N), which is a rough estimate here

because I didnt let N vary and I didnt include what

the value of m is. So you have to polynomials evaluated

in the different bases b with the same coefficients c_i.

The number of real c.c. pairs (x-\alpha_i)*(x-\alpha_j)^\star

= integer or (x-\alpha_i) an integer span the prime factors.

Arent there theorems that tell you when polynomials have

non-integer root pairs so that you can make a statement

towards their construction. I think there are. There

should be large classes of integers N of the same degree

and with the same coefficients c_j but with varying base;

I think these can be constructed in ln^3 N_{max} time with

N_{max} being the largest integer considered.

The construction I refer to is given in Fast Factoring

by Dr. Gordon Chalmers. - --- gordon_as_number <gordon_as_number@...> wrote:
> Only a comment.

First you need to specify the ring in which you're working.

>

> If you want to prove a polynomials P(x) is prime,

You do not do so.

One cannot assume Z[x] given that many of the examples we've seen have

been instead in Q[x].

> isnt enough to determine the assoiated roots.

Bases ought to be irrelevant. If they're not, you're doing something wrong.

>

> E.g. P(x)=x^2+2

>

> The number N is P(x) which can base expanded in

> some base b as

>

> N=\sum_{i=0}^m c_i b^i .

> Determining the roots and the coefficients I claim

Order(funtion(N)) only makes sense if you let N vary.

> is order \ln^3(N), which is a rough estimate here

> because I didnt let N vary

You've reached that level of illucidity again...

> and I didnt include what

Yup - illucid. Polynomials can be evaluated, but their value is

> the value of m is. So you have to polynomials evaluated

> in the different bases b

independent of any base.

> with the same coefficients c_i.

The c_i, as you introduced them, are not coefficients.

> The number of real c.c. pairs (x-\alpha_i)*(x-\alpha_j)^\star

With no firm foundation to build on, you've completely lost me now.

> = integer or (x-\alpha_i) an integer span the prime factors.

No point in continuing.

Phil

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